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If $X$ is a separable topological space, and $V$ is some dense subspace of it, is $V$ necessarily separable?

Davide Giraudo
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aellab
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  • Is $X$ a metric space? – Hamou Aug 09 '14 at 14:38
  • Any subspace of a separable metric space is again separable (because in metric spaces, second-countable <=> separable, and second-countability is hereditary) – Pedro Aug 09 '14 at 14:39
  • Oops! I didn't see the density requirement! :-( But I'm sure there are duplicates too. :-) – Asaf Karagila Aug 09 '14 at 14:52
  • Every open subset of a separable space is separable. It fails for closed sets (e.g. the antidiagonal in the Sorgenfrey plane). It also fails for dense sets, see the examples. – Henno Brandsma Aug 10 '14 at 08:11
  • @HennoBrandsma Are you talking about general topological spaces in your comment? Otherwise, as Pedro said, any subset, equipped with the inherited metric, of a separable metric space is separable. – 0xbadf00d Jun 24 '17 at 21:08
  • @0xbadf00d the open subspace part holds for all spaces. – Henno Brandsma Jun 24 '17 at 21:10

2 Answers2

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No. Let $X=\omega_1+1$, where $\omega_1$ is the first uncountable ordinal. Let $U\subseteq X$ be open iff $U$ is empty or if it contains some uncountable tail, i.e. if $\{\alpha;\alpha>\beta\}\subseteq U$ for some $\beta<\omega_1$. One easily checks that this is a topology.

Now observe that the point $\omega_1$ is in every nonempty open set and therefore $X$ is the closure of this single point and thus separable. In fact, it's not hard to see that a subspace $Y$ is dense in $X$ iff $Y$ is unbounded. So consider the subspace $Y=\omega_1\subseteq X$. By our observation $Y$ is dense in $X$ but is not itself separable since no countable set can be unbounded in $Y$.

Miha Habič
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A classical example of this: let $X = \prod_{i \in I} [0,1]_i$, where $I$ has size $|\mathbb{R}|$ and all $[0,1]_i$ are copies of $[0,1]$. Then $X$ is a compact separable set (Hewitt-Marczewski-Pondizcery theorem), but the sigma product $Y = \Sigma_{i \in I} [0,1]$ (all $(x_i) \in X$ with such that all but countably many coordinates are non-zero) is dense in $X$, but not separable, as every countable set in $Y$ only depends on countably many coordinates, and so cannot be dense. $Y$ is also countably compact (and not compact, of course), for similar reasons.

Henno Brandsma
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